If you multiply through you are left with
\[\Large 1=A(s+4)^2+B(s+4)+Cs \]

anonymous

5 years ago

I thought you are supposed to multiply A by the values under B and C?

anonymous

5 years ago

I made a mistake above, but no you want to solve for the coefficients so you might get an easier expression

anonymous

5 years ago

\[ \Large 1=A(s+4)^2+Bs(s+4)+Cs\]

anonymous

5 years ago

better now it seems

anonymous

5 years ago

wouldnt you get A=0, B=0 and C=-1/4

anonymous

5 years ago

Sorry for the mistake again, I messed up the above form, so if you solved for that your answer will be wrong.

anonymous

5 years ago

\[\Large 1=(A+B)s^2 +(8A+4B+C)s+16A \]

anonymous

5 years ago

could you please explain how you go that

anonymous

5 years ago

\[A+B=0 \\ 8A+4B+C=0
\\
16A=0 \]

anonymous

5 years ago

Yes, I multiplied the above form by s(s+4)^2 (make sure you cancel on the right hand side right) and then you factor likewise terms. When you did that, you can set both sides of the equation equal, they are equal if and only if their coefficients match.

anonymous

5 years ago

\[\Large \frac{1}{s(s+4)^2} = \frac{A}{s}+ \frac{B}{s+4}+ \frac{C}{(s+4)^2} \]
General setup of partial fraction decomposition, multiply through by \(s(s+4)^2\), it will disappear entirely on the left hand side of the equation and piecewise on the righthandside of the equation.

anonymous

5 years ago

Then expand the righthand side and factor likewise terms, set the coefficients equal to the coefficients on the lefthand side to make a valid equation out of it LHS=RHS, in this case a lot of the coefficients have to be zero, because there is only 1 on the LHS.

anonymous

5 years ago

you will end up with a 3x3 system of linear equations.
\[A+B=0 \\ 8A+4B+C=0
\\
16A=1 \]
*corrected the last line*

anonymous

5 years ago

okay so far?

anonymous

5 years ago

sorry im still confused on how you got the equation

anonymous

5 years ago

can you please show me by the work?

anonymous

5 years ago

The idea of partial fraction decomposition is that you find a different way of writing the quotient, so in the end there will be an easier form for it which is especially useful in integral calculus and for LaPlace Transformations.
The idea is that you completely factor the denominator and then you split it up, so each new quotient becomes its own denominator

anonymous

5 years ago

|dw:1344895643951:dw|

anonymous

5 years ago

i understand that part but after that I get confuesd on how you distributed the variables

anonymous

5 years ago

\[ \Large 1=(A+B)s^2 +(8A+4B+C)s+16A \] This part?

anonymous

5 years ago

yea, I'm not sure how you got that

anonymous

5 years ago

I multiplied both sides by s(s+4)^2 so the denominator will completely disappear on the LHS, on the right hand side I have:
\[\Large 1=A(s+4)^2+Bs(s+4)+Cs \]

anonymous

5 years ago

you want to solve for A, B, C in the end, but first you distribute all the binomials out and then factor likewise powers together to make a suitable system of equations.

anonymous

5 years ago

\[\Large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]

anonymous

5 years ago

now read the equation like this, on the righthand side you have a quadratic polynomial, but on the left hand side, you can understand it like this:
\[0s^2 +0s+\large 1= As^2+8As+16A + Bs^2 +4Bs + Cs \]

anonymous

5 years ago

okay?

anonymous

5 years ago

so in order that the lefthand side is equal to the righthand side, the coefficients A,B,C on the righthand side have to match the coefficients of the same powers of s on the lefthand side.

anonymous

5 years ago

\[0s^2+0s+1=(A+B)^2+(8A+4B+C)s+16A \]

anonymous

5 years ago

im sorry but i dont understand the LHS

anonymous

5 years ago

I mean RHS

anonymous

5 years ago

\[ 0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A\]

anonymous

5 years ago

where did the numbers come in from

anonymous

5 years ago

just a correction, was mistyping it.

anonymous

5 years ago

and why is A added to B

anonymous

5 years ago

Well the numbers come from because you expand a polynomial, namely (s+4)^2

anonymous

5 years ago

And \[ As^2+Bs^2 = (A+B)s^2\]

anonymous

5 years ago

just normal algebra.

anonymous

5 years ago

ok so (s+4)^2=s^2+8s+16 but how do we incorproate that into the vaues for a,b and c

anonymous

5 years ago

\[\Large 1=A(s+4)^2+Bs(s+4)+Cs \]
Could you follow to this step? You see that A, B, and C gets multiplied with all that stuff. A with the first entire term, B with the second entire term, and C with the third.
To match the equation we have to write the LHS different and then factor terms together, not only the first expression will have an x^2, also the second one with the Bs in front of it!

yes, in this case I suggest it, there are methods called cover up etc which can be applied to get at least some of the solutions easily, but I wouldn't recommend it.

anonymous

5 years ago

For instance if you set \( s=-4\) in the equation above, the first terms will all go to zero and you are left with
\[ \Large 1=-4C\]

anonymous

5 years ago

but I don't recommend doing this, because this isn't an easy partial fraction decomposition.

anonymous

5 years ago

so then you get As^2+8As+16A+Bs^2+4Bs+Cs

anonymous

5 years ago

yes, that's what you get. If you read your equation careful enough you will see that there are multiple s^2 and s in it, you want to collect them and then match them to the left hand side, this is a valid technique.

anonymous

5 years ago

ok now I understand that part! thank you! after that you solve for the variable right

anonymous

5 years ago

\[0s^2+0s+1=(A+B)s^2+(8A+4B+C)s+16A \]
See, the right hand side, you multiply A+B times s^2, if A+B are 0, then the right hand side becomes 0s^2, so it is equal to the left hand side and we are happy.